Question

The sum of first $n$ terms of the $A.P$. $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+........$ is ......

• A. $\frac{n(n+1)}{2}$
• B. $2n(n+1)$
• C. $\frac{n(n+1)}{\sqrt{2}}$
• D. $1$

Answer 1

Answer: (C)

$\frac{n(n+1)}{\sqrt{2}}$

The series can be rewritten as $\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+.....$
So the nth sum $S_n=\sqrt{2}\times (1+2+3+.....+n)$

$=\frac{\sqrt{2}\times n\times (n+1)}{2}$

$=\frac{n(n+1)}{\sqrt{2}}$